• # question_answer If the roots of the equations ${{x}^{2}}-bx+c=0$ and ${{x}^{2}}-cx+b=0$ differ by the same quantity, then  $b+c$ is equal to                            [BIT Ranchi 1969; MP PET 1993] A) 4 B) 1 C) 0 D) -4

Let the roots are $\alpha ,\beta$ of  ${{x}^{2}}-bx+c=0$and ${\alpha }',{\beta }'$be roots of ${{x}^{2}}-cx+b=0$ Now $\alpha -\beta =\sqrt{{{(\alpha +\beta )}^{2}}-4\alpha \beta }=\sqrt{{{b}^{2}}-4c}$ .....(i) and $\alpha '-\beta '=\sqrt{{{(\alpha '+\beta ')}^{2}}-4\alpha '\beta '}=\sqrt{{{c}^{2}}-4b}$ .....(ii) But $\alpha -\beta =\alpha '-\beta '$ Þ $\sqrt{{{b}^{2}}-4c}=\sqrt{{{c}^{2}}-4b}\,\,\Rightarrow \,\,{{b}^{2}}-4c={{c}^{2}}-4b$ Þ ${{b}^{2}}-{{c}^{2}}=4c-4b$ Þ $(b+c)(b-c)=4(c-b)$Þ$b+c=-4$